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Svetradugi [14.3K]
3 years ago
5

I need help understanding this problem I just joined my new school and I don’t understand this can someone help me?

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
8 0

Answer:150

Step-by-step explanation:p

100

×

18

=

27

100

18

×

p

100

×

18

=

100

18

×

27

100

18

×

p

100

×

18

=

2700

18

p

=

2700

18

p

=

150

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Find the values of m and b that make the following function differentiable.
dimaraw [331]
Now, bearing in mind that, for a graph, to be "differentiable", means a "smooth transition", not an abrupt edge or a "cusp", the first subfunction of the piece-wise is just a quadratic, the second subfunction is a linear, well, so the linear needs to hit the quadratic "smoothly", not make an abrupt edge, for that, the linear must continue the quadratic from 2 onwards, since 2 is the range edge.

now, what is the piece-wise when x = 2? well, f(2) = x² or (2)² or 4

that means, the linear needs to hit it at that point, to make it so, let's make it the slope of the linear, the same as the quadratic's.

what's the quadratic's slope? well, simple enough

\bf \left. \cfrac{dy}{dx}=2x \right|_{x=2}\implies 4\impliedby \textit{this means }&#10;\begin{array}{llll}&#10;mx+b\\&#10;\uparrow \\&#10;4&#10;\end{array}

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\bf 4 = 4(2) + b\implies 4=8+b\implies 4-8=b\implies \boxed{-4=b}&#10;\\\\\\&#10;mx+b\implies 4x-4\\\\&#10;-------------------------------\\\\&#10;f(x)=&#10;\begin{cases}&#10;x^2&x\le 2\\&#10;4x-4&x\ \textgreater \ 2&#10;\end{cases}

check the picture below.

4 0
3 years ago
⦁ Find the sum of the series . 15 sigma n=1, (2n-1) Show your work
sergejj [24]

Answer:

225

Step-by-step explanation:

The sum of this series is given by the formula:

S_n=\frac{n}{2}[2a+(n-1)d]

Where

S_n is the sum

a is the first term (we get this by plugging in n = 1 into "2n-1", we have 2(1) - 1 = 1)

n is the number of terms (the sum is defined for n = 1 to n = 15, so 15 terms)

d is the common different (the difference is successive terms. Here, 2nd term would be 2(2) - 1 = 3, and first term was 1, so d = 3 -1 = 2)

<em>plugging in the info, we will get the sum:</em>

<em>S_n=\frac{n}{2}[2a+(n-1)d]\\S_{15}=\frac{15}{2}[2(1)+(15-1)(2)]\\=225</em>

<em>The sum is 225</em>

4 0
3 years ago
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